Riddles!

[NOTS3W]

A new one. Three men go fishing, and hire a boat. The boatlessor charges them for $30, so each man pays $10. While the men are walking to the landing-stage, the lessor remembers that the price is $25, so he sends his boy with $5 restitution. The boy decides that dividing $5 in three men is difficult, so he puts $2 in his pocket, and gives each man $1.

Now each man has paid $9, $27 in total. The boy has $2, that makes $29. Where is the remaining dollar?

There is no remaining dollar.

The men paid $27. $2 is in the boy's pocket and $25 is with the lessor.

[mark]

My solution is incorrect. I don't know why I assumed there was a one hour window for the prisoners to die. It was only stated that they would die on the ninth day.

As per Idontwantspams request (below) I have removed my question. It wasn't really a puzzle question anyway. I might bring it up in a separate thread.

@Mijzelf (two posts below) I understand 2 to the eighth power and you get 256. I understand the one bit of info from dieing or not. I don't understand why or how you are coming up with that formula for that application. I am not following what you are saying.After resolving that my solution was incorrect, I went and looked at NOTS3W's correct answer. I puzzled over his answer for 15-20 minutes and doodled things out on paper. Once I had done that, I understood his answer and Mijzelf's response about 2 to the eighth power. I finally grasped the scenario. NOTS3W's answer was a puzzle in itself for me.

Second paragraph I understand.

Third paragraph: You can't do that because of time constraints. In order to have your second stage, you would need to know who was dead to create your next set of groups.

I re-did the picture because it was a bit muddled and I don't think I explained what I did very clearly. Yes, you could leave someone out of the first hour's test but never mind.

In four consecutive hours you follow the pattern I laid out. First and second hours are pretty much self explanatory. Third hour shows individual columns with two numbers below each. Each number is the number of the prisoner who will drink from that column in that hour. Fourth hour shows three numbers to the right of each row and each number is the number each prisoner who will drink from the bottles in that row. At the time of reckoning someone will die in the first hour (prisoner 8 in my example). In the following hour, the dead prisoner has a red square around his number. It will follow on so through the fourth hour. I can't see how it wouldn't work. Clarified picture

@Idontwantspam (below)"OK, Not S3w got it right. No more attempting to answer some other way." Sorry, I have eight prisoners stuck in my head. I think I have finished chewing on this bone. I hope so.

Edited October 6, 2007 by DL

[Idontwantspam]

Whoa whoa whoa!!

OK, Not S3w got it right. No more attempting to answer some other way. He won. It's done.

Also, you are ONLY permitted to post a new riddle IF YOU solved the previous one, or if no-one could answer the previous one, and they answered it and personally gave you permission to post the next one. So all you jumping the gun, go edit those posts, and hopefully some moderator can clean this up a bit for us and remove the riddles not yet supposed to be available. I believe Mijzelf answered Not S3w's riddle, who answered mine. Whoever solves Mijzelf's riddle gets to post the next one...

So, spacesurfer and DL need to remove their riddles for now, please. Thanks.

[Mijzelf]

@Idontwantspam: NOTS3W already did:

There is no remaining dollar.

The men paid $27. $2 is in the boy's pocket and $25 is with the lessor.

@DL: I've been thinking about this. From point of information, the 'simple' approach is: Each man dies or not, containing 1 bit of information. So 8 men can find a poisoned bottle in 2^8=256 bottles.

Given you've got enough time, on man could do this. Drink one bottle, wait, if he didn't die, take the next bottle. Etc. The information is now not the dieing itself (he *will* die), but the exact moment he dies.

Your approach is a combination of these two, because you don't have enough time. AFAICS you can divide the original 1000 bottles in 9 groups (no man dieing marks the 9th group). Then you've got at least 7 men remaining, so you can divide the selected group in 8. etc. With four iteration steps you should be able to find the bottle in 9*8*7*6=3024 bottles. (Actually a few more, the 9th group can be bigger than the others, because you will have 8 men left to investigate this group, and this recurses also).

2nd answer to DL:

Third paragraph: You can't do that because of time constraints.

Sorry I was not more clear. I forgot to mention that in the second case the time can be reused. Give the man 1 bottle, one minute later the second, etc. After nine days and a 1000 minutes you can know which bottle was poisoned, *if* the poison is accurate enough to let a man die in 9 days +/- 30 seconds.

(Actually, you have a whole day to spend, this gives you 86 seconds for each test). Because this constraint is unlikely, you'll have to use use a longer period for each test. So I tried to reacon what is the largest possible number of bottles which could been tested given 8 man in 4 shifts, as you suggested, which is at least 3024. And this case you can also reuse the time, by using a smart scheme, as you did.

So yes, you are right. Your scheme will find the bottle. (if the poison is accurate enough for 9 days +/- 3 hours (as you can use a period of 6 hours between the tests)). Small gloss: your scheme is a waste of prisoners. In your 'clarified picture' all prisoners are dead after the 4th test, while 1 dead man per shift is enough.

Edited October 6, 2007 by Mijzelf